Respuesta :
Answer:
36.9 g is the mass of octane that is left over
Explanation:
We state the reaction by the information given.
Octane as a reactant → C₈H₁₈
Oxygen as the other reactant → O₂
We have both masses, so we can predict the limiting reactant
We convert the mass to moles:
94 g / 114g/mol = 0.824 moles of octane
200 g / 32 g/mol = 6.25 moles of O₂
The reaction is:
2C₈H₁₈(l) + 25O₂(g) → 16CO₂(g) + 18H₂O(g)
2 moles of octane need 25 moles of O₂ to react
Then, 0.824 moles of octane may react with (0.824 . 25) /2 = 10.3 moles O₂
We have 6.25 moles and we need 10.3, then the O₂ is the limiting reagent
Octane is the excess reactant:
25 moles of O₂ may react with 2 moles of octane
Therefore, 6.25 moles of O₂ will react with (6.25 . 2) / 25 = 0.500 moles
We have 0.824 moles of octane and we only need 0.500, therefore
(0.824-0.500) = 0.324 moles are left over by the reaction.
We convert the moles to mass → 0.324 mol . 114g / 1 mol = 36.9 g
Answer:
There would remain 36.9 grams of octane
Explanation:
Step 1: Data given
Mass of octane = 94.0 grams
Molar mass of octane = 114.23 g/mol
Mass of oxygen = 200 grams
Molar mass of oxygen = 32.0 g/mol
Step 2: The balanced equation
2 C8H18 + 25 O2→ 16 CO2 + 18 H2O
Step 3: Calculate moles
Moles = mass / molar mass
Moles octane = 94.0 grams / 114.23 g/mol
Moles octane = 0.823 moles
Moles oxygen = 200 grams / 32.0 g/mol
Moles oxygen = 6.25 moles
Step 4: Calculate the limiting reactant
For 2 moles octane we need 25 moles oxygen to produce 16 moles CO2 and 18 moles H2O
The limiting reactant is oxygen. It will completely be consumed (6.25 moles). Octane is in excess. There will react 6.25 / 12.5 = 0.5 moles
There will remain 0.823 - 0.500 = 0.323 moles
Step 5: Calculate mass of octane
Mass octane = Moles octane * molar mass octane
Mass octane = 0.323 moles * 114.23 g/mol
Mass octane = 36.9 grams
There would remain 36.9 grams of octane
